TANGENT, LINEAR APPROXIMATION,TAYLOR APPROXIMATION

Robert Marık

January 21, 2006

Contents

1 Tangent 2

2 Linear approximation 9

3 Higer order approximation 15// / . .. c©Robert Marık, 2006 ×

1 Tangent

The derivative of the function f (x) at a point x0 is the slope of the tangent tothe graph of the funciton f at the point x0. Having the slope f ′(x0) and onepoint of the line (the point [x0, f (x0)]), it is easy to write the point-slope formof the tangent as

y = f ′(x0)(x− x0)+ f (x0).

// / . .. c©Robert Marık, 2006 ×

Find tangent to the graph of y =1

√1− x

at the point x0 = 0.

general formula: y = y′(x0) · (x− x0)+ y(x0)

y(x0) =1

√1−0

= 1

y′(x) =(

(1− x)−12

)′=−

12(1− x)−

32 (−1)

∣∣∣∣∣x=x0

=12

tangent: y =12· (x−0)+1 = 1+

12

x

The best linear approximation for small x:1

√1− x

≈ 1+12

x

// / . .. c©Robert Marık, 2006 ×

Find tangent to the graph of y =1

√1− x

at the point x0 = 0.

general formula: y = y′(x0) · (x− x0)+ y(x0)

y(x0) =1

√1−0

= 1

y′(x) =(

(1− x)−12

)′=−

12(1− x)−

32 (−1)

∣∣∣∣∣x=x0

=12

tangent: y =12· (x−0)+1 = 1+

12

x

The best linear approximation for small x:1

√1− x

≈ 1+12

x

We start with the general formula for the tangent.

// / . .. c©Robert Marık, 2006 ×

Find tangent to the graph of y =1

√1− x

at the point x0 = 0.

general formula: y = y′(x0) · (x− x0)+ y(x0)

y(x0) =1

√1−0

= 1

y′(x) =(

(1− x)−12

)′=−

12(1− x)−

32 (−1)

∣∣∣∣∣x=x0

=12

tangent: y =12· (x−0)+1 = 1+

12

x

The best linear approximation for small x:1

√1− x

≈ 1+12

x

We evaluate the function at the point undder consideration.

// / . .. c©Robert Marık, 2006 ×

Find tangent to the graph of y =1

√1− x

at the point x0 = 0.

general formula: y = y′(x0) · (x− x0)+ y(x0)

y(x0) =1

√1−0

= 1

y′(x) =(

(1− x)−12

)′=−

12(1− x)−

32 (−1)

∣∣∣∣∣x=x0

=12

tangent: y =12· (x−0)+1 = 1+

12

x

The best linear approximation for small x:1

√1− x

≈ 1+12

x

We differentiate.

// / . .. c©Robert Marık, 2006 ×

Find tangent to the graph of y =1

√1− x

at the point x0 = 0.

general formula: y = y′(x0) · (x− x0)+ y(x0)

y(x0) =1

√1−0

= 1

y′(x) =(

(1− x)−12

)′=−

12(1− x)−

32 (−1)

∣∣∣∣∣x=x0

=12

tangent: y =12· (x−0)+1 = 1+

12

x

The best linear approximation for small x:1

√1− x

≈ 1+12

x

We evaluate the derivative at the point under consideration.

// / . .. c©Robert Marık, 2006 ×

Find tangent to the graph of y =1

√1− x

at the point x0 = 0.

general formula: y = y′(x0) · (x− x0)+ y(x0)

y(x0) =1

√1−0

= 1

y′(x) =(

(1− x)−12

)′=−

12(1− x)−

32 (−1)

∣∣∣∣∣x=x0

=12

tangent: y =12· (x−0)+1 = 1+

12

x

The best linear approximation for small x:1

√1− x

≈ 1+12

x

We use the general formula and find the tangent, which is the best locallinear approxiamtion for the function.

// / . .. c©Robert Marık, 2006 ×

2 Linear approximation

The tangent is the best linear approximation to the function. Thisapproximation can be used to replace some complicated and inconvenientformulas by simpler ones.

// / . .. c©Robert Marık, 2006 ×

1√

1− x≈ 1+

12

x

E = mc2 = m0c2 1√

1−(v

c

)2

≈ m0c2(

1+12

(vc

)2)

=

rest massenergy︷ ︸︸ ︷

m0c2 +12

m0v2

︸ ︷︷ ︸

kinetic energy

The function can be replaced by its tangent for small x.

// / . .. c©Robert Marık, 2006 ×

1√

1− x≈ 1+

12

x

E = mc2 = m0c2 1√

1−(v

c

)2

≈ m0c2(

1+12

(vc

)2)

=

rest massenergy︷ ︸︸ ︷

m0c2 +12

m0v2

︸ ︷︷ ︸

kinetic energy

We start with Einstein formula for total energy of a moving body.

// / . .. c©Robert Marık, 2006 ×

1√

1− x≈ 1+

12

x

E = mc2 = m0c2 1√

1−(v

c

)2

≈ m0c2(

1+12

(vc

)2)

=

rest massenergy︷ ︸︸ ︷

m0c2 +12

m0v2

︸ ︷︷ ︸

kinetic energy

The mass is an increasing function of the velocity, as the formula shows.

// / . .. c©Robert Marık, 2006 ×

1√

1− x≈ 1+

12

x

E = mc2 = m0c2 1√

1−(v

c

)2

≈ m0c2(

1+12

(vc

)2)

=

rest massenergy︷ ︸︸ ︷

m0c2 +12

m0v2

︸ ︷︷ ︸

kinetic energy

The formula contains our function for x =(v

c

)2. We replace it by its

tangent. This can be done if the velocity is many times smaller than c, thevelocity of the light in vacuum..

// / . .. c©Robert Marık, 2006 ×

1√

1− x≈ 1+

12

x

E = mc2 = m0c2 1√

1−(v

c

)2

≈ m0c2(

1+12

(vc

)2)

=

rest massenergy︷ ︸︸ ︷

m0c2 +12

m0v2

︸ ︷︷ ︸

kinetic energy

After some algebraic modifications we get the part of the rest mass energy(not connected with any motion) and the part which includes the velocityand thus connected with the motion. This part corresponds with the formulafor kinetic energy known from Newtonian mechanics.

// / . .. c©Robert Marık, 2006 ×

3 Higer order approximation

The linear approximation is usually limited to a narrow neigborhood only. Ifthe range of variables is greater than this neighborhood and the linearapproximation does not give satisfactory results, it is possible to use higherorder approximation by higher order Taylor polynomial.

// / . .. c©Robert Marık, 2006 ×

y =1

√1− x

y′ =12(1− x)−3/2

y′′ =34(1− x)−5/2

y′′′ =158

(1− x)−7/2

y(0) = 1

y′(0) =12

y′′(0) =34

y′′′(0) =158

1√

1− x≈ 1+

12

x+34

12!

x2 +158

13!

x3 + · · ·

≈ 1+12

x+38

x2 +516

x3 + · · ·

1√

1− x≈ 1+

12

x+38

x2 +516

x3 + · · ·

E = m0c2 1√

1−(v

c

)2≈

rest massenergy︷︸︸︷

m0c2 +12

m0v2

︸ ︷︷ ︸

classicalkinetic energy

+

+38

m0v4

c2︸ ︷︷ ︸

the 1-st relativisticcorrection

+5

16m0

v6

c4︸ ︷︷ ︸

the 2-nd relativisticcorrection

+ · · ·

We find higher order approximation for the Einstein’s formula from thepreceeding chapter. We differentiate up to third order derivative.

// / . .. c©Robert Marık, 2006 ×

y =1

√1− x

y′ =12(1− x)−3/2

y′′ =34(1− x)−5/2

y′′′ =158

(1− x)−7/2

y(0) = 1

y′(0) =12

y′′(0) =34

y′′′(0) =158

1√

1− x≈ 1+

12

x+34

12!

x2 +158

13!

x3 + · · ·

≈ 1+12

x+38

x2 +516

x3 + · · ·

1√

1− x≈ 1+

12

x+38

x2 +516

x3 + · · ·

E = m0c2 1√

1−(v

c

)2≈

rest massenergy︷︸︸︷

m0c2 +12

m0v2

︸ ︷︷ ︸

classicalkinetic energy

+

+38

m0v4

c2︸ ︷︷ ︸

the 1-st relativisticcorrection

+5

16m0

v6

c4︸ ︷︷ ︸

the 2-nd relativisticcorrection

+ · · ·

We evaluate the function and its derivatives at zero.

// / . .. c©Robert Marık, 2006 ×

y =1

√1− x

y′ =12(1− x)−3/2

y′′ =34(1− x)−5/2

y′′′ =158

(1− x)−7/2

y(0) = 1

y′(0) =12

y′′(0) =34

y′′′(0) =158

1√

1− x≈ 1+

12

x+34

12!

x2 +158

13!

x3 + · · ·

≈ 1+12

x+38

x2 +516

x3 + · · ·

1√

1− x≈ 1+

12

x+38

x2 +516

x3 + · · ·

E = m0c2 1√

1−(v

c

)2≈

rest massenergy︷︸︸︷

m0c2 +12

m0v2

︸ ︷︷ ︸

classicalkinetic energy

+

+38

m0v4

c2︸ ︷︷ ︸

the 1-st relativisticcorrection

+5

16m0

v6

c4︸ ︷︷ ︸

the 2-nd relativisticcorrection

+ · · ·

We write the Taylor polynomial.

// / . .. c©Robert Marık, 2006 ×

y =1

√1− x

y′ =12(1− x)−3/2

y′′ =34(1− x)−5/2

y′′′ =158

(1− x)−7/2

y(0) = 1

y′(0) =12

y′′(0) =34

y′′′(0) =158

1√

1− x≈ 1+

12

x+34

12!

x2 +158

13!

x3 + · · ·

≈ 1+12

x+38

x2 +516

x3 + · · ·

1√

1− x≈ 1+

12

x+38

x2 +516

x3 + · · ·

E = m0c2 1√

1−(v

c

)2≈

rest massenergy︷︸︸︷

m0c2 +12

m0v2

︸ ︷︷ ︸

classicalkinetic energy

+

+38

m0v4

c2︸ ︷︷ ︸

the 1-st relativisticcorrection

+5

16m0

v6

c4︸ ︷︷ ︸

the 2-nd relativisticcorrection

+ · · ·

We simplify.

// / . .. c©Robert Marık, 2006 ×

1√

1− x≈ 1+

12

x+38

x2 +516

x3 + · · ·

E = m0c2 1√

1−(v

c

)2≈

rest massenergy︷︸︸︷

m0c2 +12

m0v2

︸ ︷︷ ︸

classicalkinetic energy

+

+38

m0v4

c2︸ ︷︷ ︸

the 1-st relativisticcorrection

+5

16m0

v6

c4︸ ︷︷ ︸

the 2-nd relativisticcorrection

+ · · ·

Higher order approximation for our function.

// / . .. c©Robert Marık, 2006 ×

1√

1− x≈ 1+

12

x+38

x2 +516

x3 + · · ·

E = m0c2 1√

1−(v

c

)2≈

rest massenergy︷︸︸︷

m0c2 +12

m0v2

︸ ︷︷ ︸

classicalkinetic energy

+

+38

m0v4

c2︸ ︷︷ ︸

the 1-st relativisticcorrection

+5

16m0

v6

c4︸ ︷︷ ︸

the 2-nd relativisticcorrection

+ · · ·

Higher order approximation of the Einstein’s formula for energy of movingobject.

// / . .. c©Robert Marık, 2006 ×

Futher reading

• http://en.wikipedia.org/wiki/Kinetic energy

• http://www.phys.unsw.edu.au/einsteinlight/jw/module5 dynamics.htm

// / . .. c©Robert Marık, 2006 ×